Answer:
8.85 x 10⁴ Nm²/C
Explanation:
d = diameter of the conducting sphere = 0.10 m
r = radius of the conducting sphere = (0.5) d = (0.5) (0.10) = 0.05 m
Area of the sphere is given as
A = 4πr²
A = 4 (3.14) (0.05)²
A = 0.0314 m²
σ = Surface charge density = 150 x 10⁻⁶ C/m²
Q = total charge enclosed
Total charge enclosed is given as
Q = σA
Q = (150 x 10⁻⁶) (0.0314)
Q = 4.7 x 10⁻⁶ C
Electric flux through one of the side is given as
[tex]\phi = \frac{Q}{6\epsilon _{o}}[/tex]
[tex]\phi = \frac{4.7\times 10^{-6}}{6(8.85\times 10^{-12})}[/tex]
[tex]\phi [/tex] = 8.85 x 10⁴ Nm²/C